# Treating the delta potential in a Schroedinger equation in 1D

It is a standard problem in quantum mechanics. For the equation

$$-\psi'' + g \delta(x) \psi = E \psi ,$$

we integrate from $$-\epsilon$$ to $$+\epsilon$$ and thus get the boundary condition

$$g \psi(0) = \psi'(0+) - \psi'(0-) .$$

Integrating again, we know the wave function is continuous at $$x=0$$,

$$\psi(0+)= \psi(0-) =\psi(0).$$

Now the conclusion is that $$\psi$$ is continuous at $$x=0$$ but $$\psi'$$ is not. The question is then, how should we interpret the product $$\delta(x) \psi$$ in the original equation? Only for an infinitely differentiable function, is the product well-defined as a distribution, right? At least this is what I see in the book (mathematics for the physical sciences) of Laurent Schwartz.

The distributional issue of multiplying a Dirac delta distribution with a non-smooth wavefunction $$\psi$$ can be avoided by rewriting the TISE as an integral equation, cf. my Phys.SE answer here.

Taken literally, $$\hat H = -\frac{1}{2}\frac{d^2}{dx^2} + \lambda \delta(x)$$ is not a genuine operator on $$L^2(\mathbb R)$$. One can see this by noting that even if you could make sense of the expression $$\delta(x)\psi(x)$$, the function

$$(\hat H\psi)(x) = -\frac{1}{2}\psi''(x) + \lambda \delta(x)\psi(x)$$

would not be square-integrable.

Consider instead the free particle on the Hilbert space $$L^2\bigg((-\infty,0)\cup(0,\infty)\bigg)$$, with inner product $$\left<\psi,\phi\right> = \lim_{a,b\rightarrow 0^+}\left[\int_{-\infty}^{-a}\overline{\psi(x)}\phi(x) dx + \int_b^\infty \overline{\psi(x)}\phi(x) dx\right]$$ The form of the Hamiltonian will simply be $$\hat H = -\frac{1}{2}\frac{d^2}{dx^2}$$, but now we must be careful about domain issues. Watch what happens when we check for Hermiticity.

$$-2\langle \psi, \hat H \phi\rangle = \lim_{a,b\rightarrow 0^+}\left[\int_{-\infty}^{-a} \overline{\psi(x)}\phi''(x) dx + \int_b^\infty \overline{\psi(x)} \phi''(x)dx\right]$$ $$=\lim_{a,b\rightarrow 0^+}\left[\overline{\psi(-a)}\phi'(-a)-\overline{\psi'(-a)}\phi(-a) - \overline{\psi(b)}\phi'(b) + \overline{\psi'(b)}\phi(b) \right] -2\left<\hat H\psi,\phi\right>$$

For the standard free particle on a line, that boundary term vanishes because if $$\psi$$ and $$\phi$$ are twice (weakly) differentiable, then they and their first derivatives must be at least continuous. On this Hilbert space however, it is possible for twice-differentiable functions to have completely different limits as $$x\rightarrow 0$$ from the left and right.

If $$\hat H$$ is to be Hermitian, then the domain consisting of twice-differentiable functions whose second derivatives are square-integrable (which is the standard domain in the $$L^2(\mathbb R)$$ case) is too big. We need to add restrictions in the form of boundary conditions to ensure that the boundary term vanishes.

You can check that the boundary conditions $$\lim_{x\rightarrow 0^+}\psi(x) = \lim_{x\rightarrow 0^-}\psi(x) = \alpha$$ $$\lim_{x\rightarrow 0^+}\psi'(x) - \lim_{x\rightarrow 0^-}\psi(x) = \lambda\alpha$$

with $$\alpha\in \mathbb C$$ and $$\lambda\in\mathbb R$$ are sufficient to kill the unwanted boundary term. However, this is precisely the set of boundary conditions which arises from the $$\delta(x)$$ potential.

To recap, we see that the free particle on the disconnected line requires a specific type of boundary condition on the domain of the Hamiltonian at the point of disconnection in order for $$\hat H$$ to be Hermitian. There are multiple choices one could make (e.g. one could require that $$\psi(x)\rightarrow 0$$ as $$x\rightarrow 0$$), but if we demand that (i) $$\psi$$ approaches the same value from the left and right, and (ii) $$\psi'$$ have a jump discontinuity equal to a real number times the aforementioned limit, then $$\hat H$$ will be Hermitian.

On the other hand, we obtain the same conditions by considering the Hamiltonian $$\hat H = -\frac{1}{2}\frac{d^2}{dx^2} + \lambda \delta(x)$$ on the Hilbert space $$L^2(\mathbb R)$$, as long as we don't ask too many questions about $$\hat H$$ at the point $$x=0$$. We can therefore consider the somewhat loose delta function Hamiltonian to be a "recipe" which gives us the same result as the more rigorous - but also somewhat more annoying - free particle on a disconnected line.